Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 4x}{x + 6} = \dfrac{-15x - 30}{x + 6}$
Explanation: Multiply both sides by $x + 6$ $ \dfrac{x^2 - 4x}{x + 6} (x + 6) = \dfrac{-15x - 30}{x + 6} (x + 6)$ $ x^2 - 4x = -15x - 30$ Subtract $-15x - 30$ from both sides: $ x^2 - 4x - (-15x - 30) = -15x - 30 - (-15x - 30)$ $ x^2 - 4x + 15x + 30 = 0$ $ x^2 + 11x + 30 = 0$ Factor the expression: $ (x + 5)(x + 6) = 0$ Therefore $x = -5$ or $x = -6$ At $x = -6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -6$, it is an extraneous solution.